If `f(x)` is a polynomial of degree `<3,` prove that `|1af(a)//(x-a)1bf(b)//(x-b)1cf(c)//(x-c)|-:|1a a^2 1bb^2 1cc^2|=(f(x))/((x-a)(x-b)(x-c))`

Explanation

we have to prove that
`|{:(1,,a,,(f(a))/((x-a))),( 1,,b,,(f(b))/((x-b))),(1,,c,,(f(c))/((x-c))):}| -: |{:(1,,a,,a^(2)),( 1,,b,,b^(2)),(1,,c,,c^(2)):}| =(f(x))/((x-a)(x-b)(x-c))`
`L.H.S.=|{:(1,,a,,(f(a))/((x-a))),( 1,,b,,(f(b))/((x-b))),(1,,c,,(f(c))/((x-c))):}| -: [(a-b)(b-c)(c-a)]`
Expanding along `C_(3)` we get
`L.H.S. =(1)/((a-b)(b-c)(c-a))xx`
`[(f(a)(c-b))/((x-a))+(f(b)(a-c))/((x-b))+(f(c)(b-a))/((x-c))]`
Now using partial fraction method on R.H.S. we get
`R.H.S. (f(x))/((x-a)(x-b)(x-c))=(A)/(x-a)+(B)/(x-b)+(C)/(x-c)` ltbr. `"(As degree of " f(x) lt 3)`
`" then " A= [(f(x))/((x-b)(x-c))]_(x=a)`
`=(f(a))/((b-a)(a-c))`
`" Similarly "=(1)/((b-a)(b-c)) " and " C = (f(c))/((c-a)(c-b))`
`:. R.H.S. =(1)/((a-b)(b-c)(c-a))xx`
`[((c-b)f(a))/((x-b))+((a-c)f(b))/((x-b))+((b-a)f(c))/((x-c))]`
Hence L.H.S. =R.H.S.`